How do you solve the following system: $2x-4y=6 , y= 3/4x+3 2$ ?

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Answer 1 · 2016-02-27T01:24:15

x=-134; y=137/2

$a) 2x-4y=6$

$b) y=3/4x+32$

You just need to replace $y$ in $a)$ by its value in $b$ :

$a) 2x-4(3/4x+32)=6$

$a) 2x-3x-128=6$

$a) -x=6+128=134$

$a) x=-134$

Now that we have the value of x we wil replace it in equation b:

$b) y=(3/4)xx(-134)+32=-201/2+32= 137/2$

How do you solve the following system: 2x-4y=6 , y= 3/4x+3 2 2x-4y=6 , y= 3/4x+3 2 ?